Re: [其他] extreme point證明

看板Math作者Vulpix (Sebastian)時間14年前 (2011/10/15 16:23)推噓8(8推 0噓 12→)

留言20則, 4人參與討論串2/2 (看更多)

※ 引述《c96a111 (拉~阿~!)》之銘言： : Prove that the extreme point of S={x | (x^t)x≦1} : are the point on its boundary. : 這題小弟我想好久了... : 真的不知道該怎去證明端點在邊界上 : 希望有高手能幫忙一下 Given x \in S i) x = 0 take y = [1 0 ... 0]^t, z = -y then x = y/2 + z/2 hence x is not an extreme point of S ii) x \in int(S), x≠0 i.e. (x^t)x < 1, x≠0 take y = x/|x|, z = 0 then x = |x|y + (1-|x|)z hence x is not an extreme point of S (∵0 < |x| < 1) iii) x \in bd(S) i.e. (x^t)x = 1 suppose that x is not an extreme point of S __ then there is a segment in S passing through x, say yz let u be the unit vector pointing inward to S along this segment ^^^^^^^^^^^^^^^^^^^^ this means that (u^t)x≦0 since x is the outward normal vector of S at x then the segment may be parametrized as x + su, where s is a real number because the segment is contained in S so [(x + su)^t](x + su)≦1 <=> 0≦s≦-2(u^t)x then y = x + au, z = x + bu, where 0≦a,b≦-2(u^t)x and x = ry + (1-r)z = x + (ra + (1-r)b)u for some r so r = b/(b-a) since 0<r<1, we have b>0, b>a, a<0 but a<0 is impossible, which is a contradiction it means that x is an extreme point Combining i, ii and iii we conclude that the extreme points of S = {x | (x^t)x≦1} are the points on its boundary -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.248.46.14

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c96a111

10/15 16:37, , 1^F

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jacky7987

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